Questions: A population P is initially 3950 and four hours later reaches the given number. Find the four-hour growth or decay factor. (Round your answers to three decimal places.) (a) 5900 (b) 9100 (c) 1725 (d) 1450

Solution Steps

To find the four-hour growth or decay factor, we use the formula for exponential growth or decay: \( P(t) = P_0 \times r^t \), where \( P_0 \) is the initial population, \( r \) is the growth/decay factor, and \( t \) is the time in hours. We solve for \( r \) by rearranging the formula to \( r = \left(\frac{P(t)}{P_0}\right)^{1/t} \).

To find the growth or decay factor over a period of time, we use the formula for exponential growth or decay: \[ P(t) = P_0 \times r^t \] where:

• \( P_0 \) is the initial population,
• \( r \) is the growth/decay factor,
• \( t \) is the time in hours.

Rearrange the formula to solve for the growth or decay factor \( r \): \[ r = \left(\frac{P(t)}{P_0}\right)^{\frac{1}{t}} \]

Using the initial population \( P_0 = 3950 \) and \( t = 4 \) hours, calculate \( r \) for each final population:

\[ r_a = \left(\frac{5900}{3950}\right)^{\frac{1}{4}} \approx 1.106 \]

\[ r_b = \left(\frac{9100}{3950}\right)^{\frac{1}{4}} \approx 1.232 \]

\[ r_c = \left(\frac{1725}{3950}\right)^{\frac{1}{4}} \approx 0.813 \]

Final Answer